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Invariant sets and solutions to a class of wave equations
Applied Mathematics and Computation 253 (2015) 369–376
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Invariant sets and solutions to a class of wave equations q Yaying Dong a,⇑, Shunli Zhang c, Shanbing Li b a b c
School of Mathematics, Northwest University, Xi’an 710069, PR China College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China Center for Nonlinear Studies, School of Mathematics, Northwest University, Xi’an 710069, PR China
a r t i c l e
i n f o
a b s t r a c t In this paper, we investigate the invariant sets and exact solutions of the (1 + 2)-dimensional wave equations. It is proven that there exists a class of solutions to the equations, which belong to the invariant set E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Wave equation Invariant set Exact solution
1. Introduction In mathematical physics, seeking exact solutions of nonlinear partial differential equations (PDEs) has been the highlights. Constructing exact solutions to nonlinear PDEs has theoretical significance and practical value in physics, engineering and mechanics, etc. Up to now, the scholars have developed various ways to solve nonlinear systems. In [1], Galaktionov introduced an extension to the scaling group for the equation
ut ¼ Kðx; ux ; uxx ; . . . ; uðkÞ Þ: It is governed by the invariance of the set S0 ¼ fu : ux ¼ FðuÞ=xg, where uðkÞ represents the kth-order derivative of u with respect to x and F is a smooth function to be determined. This approach is available to solve the nonlinear evolution equations of the KdV-type[2], and it has also been used to study the second and fourth-order nonlinear evolution equations [1]. In [3], Qu and Estevez extended the scaling group to a more general form, which is governed by the invariant set
S1 ¼
u : ux ¼
Z 1 FðuÞ þ FðuÞ expðn 1Þ x
u
1 dz : FðzÞ
This method has been used successfully to construct exact solutions to a large amount of evolution equations (see [3,4]). Moreover, in [5,6], Qu and Zhu further extended the Galaktionov’s method to study the generalized thin film equation and the (1 + 2)-dimensional reaction–diffusion equations by introducing the invariant set E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg. There have been many works contributed to constructing exact solutions of the nonlinear diffusion equation (see [7–9]). Symmetry-related methods are valid to construct exact solutions of nonlinear diffusion equations. Local symmetry-related methods including nonclassical symmetry method due to Bluman and Cole [9], Lie point symmetry method due to Lie [10], and conditional Lie Bäcklund symmetry method due to Zhdanov [8]. Another important approach is the ansatz-based method, which has been used to obtain exact solutions of the following equation
ut ¼ DGðuÞ þ HðuÞ; q
The project is supported by the National Natural Science Foundation of China (No. 11371293).
⇑ Corresponding author.
E-mail address: [email protected] (Y. Dong). http://dx.doi.org/10.1016/j.amc.2014.12.093 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
370
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
even for higher-dimensional case. It is noted that the symmetry-related methods have been used successfully to study the symmetry properties and the solutions of nonlinear PDEs (see [11–15]). In present paper, we extend Galaktionov’s approach [1,2] to study a class of wave equations and their solutions, which are governed by the invariant set E0 . Note that the invariant set is a natural generalization of S0 in [1] for the (1 + 1)-dimensional case. Consider the (1 + 2)-dimensional wave equation
utt ¼ AðuÞuxx þ BðuÞuyy þ CðuÞu2x þ DðuÞu2y þ EðuÞux þ PðuÞuy þ Q ðuÞ;
ð1:1Þ
where AðuÞ; BðuÞ; CðuÞ; DðuÞ; EðuÞ; PðuÞ and Q ðuÞ are smooth functions of u. Eq. (1.1) has a wide range of physical applications in fluid dynamics, plasma and elastic media, etc. Now, we introduce the invariant set E0 which is governed by
E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg; where v ðx; yÞ is a smooth function of x and y; FðuÞ is a smooth function of u, which is to be determined from the invariant condition
uðx; y; 0Þ 2 E0 ) uðx; y; tÞ 2 E0 ; t 2 ð0; 1: For u 2 E0 , we can obtain the solutions of Eq. (1.1), which are given by
Z
u
1 dz ¼ v ðx; yÞ þ hðtÞ: FðzÞ
In the invariant set E0 , we derive the following formulas 0
00
02
ut ¼ h F;
utt ¼ h F þ h FF 0 ;
ux ¼ v x F;
uxx ¼ v xx F þ v 2x FF 0 ;
uy ¼ v y F;
uyy ¼ v yy F þ v 2y FF 0 :
Substituting them into Eq. (1.1), we have 00
02
h þ h F 0 ¼ Av xx þ Bv yy þ ðAF 0 þ CFÞv 2x þ ðBF 0 þ DFÞv 2y þ Ev x þ Pv y þ
Q : F
ð1:2Þ
Differentiating (1.2) with respect to x and y respectively, we have
0 Q 02 h F 00 v x F ¼ Av xxx þ Bv yyx þ A0 v xx þ B0 v yy þ v x F þ ðAF 0 þ CFÞ0 F v 3x F 0
þ 2ðAF 0 þ CFÞv x v xx þ ðBF 0 þ DFÞ v x F v 2y þ 2ðBF 0 þ DFÞv y v yx
ð1:3Þ
þ Ev xx þ E0 v 2x F þ P0 v x v y F þ Pv yx ; 0 Q 02 h F 00 v y F ¼ Av xxy þ Bv yyy þ A0 v xx þ B0 v yy þ E0 v x þ P0 v y þ v yF F 0
0
þ ðAF 0 þ CFÞ F v y v 2x þ 2ðAF 0 þ CFÞv x v xy þ ðBF 0 þ DFÞ F v 3y
ð1:4Þ
þ 2ðBF 0 þ DFÞv y v yy þ Ev xy þ Pv yy : By means of Eqs. (1.3) and (1.4), it is very difficult to determine the coefficient functions in Eq. (1.1). To illustrate our method, in Section 2, we only choose certain v ðx; yÞ and calculate the exact solutions of Eq. (1.1). 2. Main results In this section, we choose certain v ðx; yÞ and construct the corresponding solutions of Eq. (1.1) to determine the invariant sets. However, we cannot determine all v ðx; yÞ such that Eq. (1.1) is invariant with respect to E0 , we can consider several special cases in the following. Case 1 v xx þ v yy ¼ 0. With respect to the solutions of the Laplace’s equation
v xx þ v yy ¼ 0; we get the following subcases.
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
Subcase 1.1 Substituting 02
h ¼
371
v ðx; yÞ ¼ ax þ by ða – 0; b – 0Þ. v ðx; yÞ ¼ ax þ by into Eq. (1.2), we obtain
1 2 Q 2 00 0 0 : a ðAF þ CFÞ þ b ðBF þ DFÞ þ aE þ bP þ h F F0
ð2:1Þ
It is noted that the left side of Eq. (2.1) does not depend on x; y, which suggests that A; B; C; D; E; P and Q satisfy
1 2 Q 2 00 0 0 ¼ c1 a ðAF þ CFÞ þ b ðBF þ DFÞ þ aE þ bP þ h F F0 and hðtÞ ¼ solution
Z
u
ð2:2Þ
pffiffiffiffiffi c1 t þ c2 , where c1 P 0 and ci denote arbitrary constants from now on. In this subcase, Eq. (1.1) has the exact
pffiffiffiffiffi 1 dz ¼ ax þ by þ c1 t þ c2 ; FðzÞ
which is the traveling wave solution. Subcase 1.2 v ¼ arctanðx=yÞ. The function v ¼ arctanðx=yÞ is also a solution of Laplace equation. In this situation, E0 becomes the set
E1 ¼
y FðuÞ; u : ux ¼ 2 x þ y2
uy ¼
x FðuÞ : x2 þ y2
If F 00 – 0, then Eqs. (1.3) and (1.4) respectively become
h
02
¼
i
1 x3 1 h x2 0 0 þ 00 6ðA BÞ þ ðBF 0 þ DFÞ F 00 2ðBF þ DFÞ 2 2 FF FF yðx2 þ y2 Þ ðx2 þ y2 Þ
1 xy 2ðA BÞ0 F þ 4ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 2 2 FF 00 ðx þ y2 Þ
i 1 h y2 ðQ =FÞ0 0 þ 00 2ðA BÞ þ ðAF 0 þ CFÞ F þ 2 2 2 FF F 00 ðx þ y Þ 2E P0 x E0 P y P x2 þ 00 00 2 þ 00 ; 00 þ 00 2 2 2 2 FF FF x þ y FF yðx þ y2 Þ F x þy F
h
02
¼
i
1 y3 1 h y2 0 0 00 6ðA BÞ ðAF 0 þ CFÞ F 00 2ðAF þ CFÞ 2 2 FF FF xðx2 þ y2 Þ ðx2 þ y2 Þ
1 xy 2ðA BÞ0 F 4ðBF 0 þ DFÞ þ 2ðAF 0 þ CFÞ 2 FF 00 ðx2 þ y2 Þ
i 1 h x2 ðQ=FÞ0 0 þ 00 2ðA BÞ þ ðBF 0 þ DFÞ F þ 2 FF F 00 ðx2 þ y2 Þ 0 0 E 2P y P E x E y2 þ 00 00 2 þ þ : 00 00 FF x þ y2 FF x2 þ y2 FF 00 xðx2 þ y2 Þ F F Differentiating Eqs. (2.3) and (2.4) with respect to x and y respectively, we obtain
0 ¼
ð2:3Þ
" 0 0 #
1 x4 BF þ DF 2x3 0 2ðBF þ DFÞ G þ F 00 00 2 2 2 2 2 2 FF FF y ðx þ y Þ yðx þ y2 Þ 6ðBF 0 þ DFÞ x2 2xy þ G0 F þ 6M þ ðG M0 F 2IÞ 00 2 2 2 2 2 FF ðx þ y2 Þ ðx þ y Þ " # 0 0 0 0 y2 ðQ =FÞ0 E 2E0 P 4P x þ ðI0 F 2MÞ þ F þ 2 F F þ 2 FF 00 FF 00 x2 þ y2 F 00 F 00 F 00 ðx2 þ y2 Þ " 0 0 0 # 2E P 0 P x2 E0 2E P0 P y þ þ þ F F F 2 ; þ x þ y2 yðx2 þ y2 Þ FF 00 F 00 FF 00 FF 00 F 00 FF 00 F 00
ð2:4Þ
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
0¼
" 0 0 #
1 y4 AF þ CF 2y3 0 2ðAF þ CFÞ H F 00 00 2 2 FF FF x2 ðx2 þ y2 Þ xðx2 þ y2 Þ
6ðAF 0 þ CFÞ y2 2xy H0 F þ þ 6N þ ðH N 0 F þ 2JÞ 00 2 2 FF ðx2 þ y2 Þ ðx2 þ y2 Þ " # 0 0 0 0 x2 ðQ=FÞ0 P 2P0 E 4E y 0 þ ðJ F þ 2NÞ þ F þ 2 F 00 þ 00 F 00 2 2 F 00 F F FF 00 FF x þ y2 ðx2 þ y2 Þ " 0 0 0 # 2P E0 E y2 P0 2P E0 E x þ 00 þ 00 þ F F 2 ; þ 00 F þ 00 00 x þ y2 xðx2 þ y2 Þ FF FF 00 FF FF 00 F F F
where
1 0 ½6ðA BÞ þ ðBF 0 þ DFÞ F; FF 00
H¼
1 0 ½6ðA BÞ ðAF 0 þ CFÞ F; FF 00
1 0 ½2ðA BÞ þ ðAF 0 þ CFÞ F; FF 00
J¼
1 0 ½2ðA BÞ þ ðBF 0 þ DFÞ F; FF 00
G¼
I¼
M¼
1 ½ðA BÞ0 F þ 2ðAF 0 þ CFÞ ðBF 0 þ DFÞ; FF 00
N¼
1 ½ðA BÞ0 F 2ðBF 0 þ DFÞ þ ðAF 0 þ CFÞ: FF 00
They provide the constraints for the coefficient functions in Eq. (1.1)
AF 0 þ CF ¼ 0;
A ¼ B;
BF 0 þ DF ¼ 0;
Q ¼ c3 F 0 ; F
0 2E P0 P þ þ F ¼ 0; FF 00 F 00 FF 00
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 ;
2
E FF 00
0
F
0 0 2E0 P 4P F þ 00 ¼ 0; FF F 00 F 00
0 0 0 E 2E P0 P F F ¼ 0: F 00 FF 00 F 00 FF 00
In this situation, Eq. (1.1) has the exact solution
Z
u
pffiffiffiffiffi 1 x þ c1 t þ c2 : dz ¼ arctan FðzÞ y
If F 00 ¼ 0, i.e., F ¼ u, similarly, we also have the above formulas. It is easy to verify that the equation m1
utt ¼ um ðuxx þ uyy Þ ku
ðu2x þ u2y Þ þ c3 ku
2k1
has the exact solution
pffiffiffiffiffi 8 > < c0 exp arctan yx þ c1 t ; k ¼ 1; u¼ h i 1 ffi > k1 : ð1 kÞ arctan x þ pffiffiffiffi ; k – 1: c1 t þ c2 y Subcase 1.3 v xx ¼ v yy ¼ 0. Since v xx ¼ v yy ¼ 0, we derive that becomes the set
v ðx; yÞ ¼ cxy þ d.
ð2:5Þ
Without loss of generality, we choose c ¼ 1 and d ¼ 0. Thus E0
E1 ¼ fu : ux ¼ yFðuÞ; uy ¼ xFðuÞg: If F 00 – 0, substituting
v xx ¼ v yy ¼ 0 into Eqs. (1.3) and (1.4) respectively, then we get
0 1 2ðBF 0 þ DFÞ x Q 1 0 0 02 h ¼ 00 ðBF 0 þ DFÞ x2 þ ðAF 0 þ CFÞ y2 þ þ E0 y þ P0 x þ þ P ; F y F Fy F
ð2:6Þ
0 1 2ðAF 0 þ CFÞ y Q 1 0 2 0 2 0 0 0 0 : E ðBF þ DFÞ x þ ðAF þ CFÞ y þ þ E y þ P x þ þ F x F Fx F 00
ð2:7Þ
02
h ¼
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
373
Differentiating Eqs. (2.6) and (2.7) with respect to x and y respectively, we obtain
" # ( ) " # 0 0 0 0 0 0 0 ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 1 ðQ=FÞ0 2 xþ 0¼ Fx y þ Fþ Fy3 þ þ Fy 00 00 00 00 00 00 y FF FF F F F F þ
0 0 0 0 0 E P P0 P 2 Fy þ Fyx þ þ F; FF 00 F 00 F 00 F 00
" # ( ) " # 0 0 0 0 0 0 0 ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ ðBF 0 þ DFÞ 2ðAF 0 þ CFÞ 1 ðQ =FÞ0 2 3 y þ Fxy þ F þ Fx þ þ Fx x F 00 F 00 F 00 F 00 FF 00 FF 00 0 0 0 0 0 E P E0 E F: þ 00 Fyx þ 00 Fx2 þ 00 þ F F F FF 00
0¼
Hence, we derive the coefficient functions in Eq. (1.1) satisfying
AF 0 þ CF ¼ 0;
BF 0 þ DF ¼ 0;
Q ¼ c4 F 0 ; F
E0 ¼ c5 F 00 ;
P 0 ¼ c6 F 00
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 :
In this situation, Eq. (1.1) has the exact solution
Z
u
pffiffiffiffiffi 1 dz ¼ xy þ d þ c1 t þ c2 : FðzÞ
If F 00 ¼ 0, i.e., F ¼ u, we also have the above formulas. k1 k1 m1 2k1 Let A ¼ B ¼ um ; F ¼ uk . We have E ¼ c5 ku ; P ¼ c6 ku ; C ¼ D ¼ ku and Q ¼ c4 ku . Thus we have shown that the equation m1
utt ¼ um ðuxx þ uyy Þ ku
k1
ðu2x þ u2y Þ þ c5 ku
ux þ c6 ku
k1
2k1
uy þ c4 ku
has the exact solution
( u¼
pffiffiffiffiffi c1 tÞ; k ¼ 1; 1 pffiffiffiffiffi ½ð1 kÞðxy þ c1 t þ c2 Þk1 ; k – 1:
c0 expðxy þ
ð2:8Þ
Case 2 v xy ¼ 0. From v xy ¼ 0, we deduce v ðx; yÞ ¼ f ðxÞ þ gðyÞ, where f ðxÞ and gðyÞ are the smooth functions of x and y respectively. In this case, E0 becomes 0
E1 ¼ fu : ux ¼ f ðxÞF; uy ¼ g 0 ðyÞFg: We now consider several subcases. Subcase 2.1 f ðxÞ ¼ ln jxj; gðyÞ ¼ ln jyj. 0 Set f ðxÞ ¼ ln jxj; gðyÞ ¼ ln jyj. Then f ðxÞ ¼ 1=x;
1 1 E1 ¼ u : ux ¼ F; uy ¼ F : x y
g 0 ðyÞ ¼ 1=y; v ðx; yÞ ¼ ln jxj þ ln jyj. Moreover, we have
Suppose that F 00 – 0. Then Eq. (1.3) becomes 02
h ¼
0 i 1 1 1 1h 1 Q E E0 1 P0 1 0 0 0 0 0 2ðA AF CFÞ ðA AF CFÞ F ðB BF DFÞ þ þ FF 00 x2 F 00 y2 F 00 F FF 00 F 00 x F 00 y
ð2:9Þ
and Eq. (1.4) becomes 02
h ¼
0 0 i 1 1 1 1h 1 Q E0 1 P P 1 0 0 0 0 0 2ðB BF DFÞ ðB BF DFÞ F ðA AF CFÞ þ þ þ : F 00 x F 00 FF 00 y FF 00 y2 F 00 x2 F 00 F
Differentiating Eqs. (2.9) and (2.10) with respect to x and y respectively, we have
" # " # 0 0 0 0 0 0 0 0
1 ðB BF 0 DFÞ 1 ðQ=FÞ0 1 E E E E0 1 P 1 0 ¼ G F 2G 3 F 2þ F F 00 þ 00 2 þ 00 F ; 00 00 00 00 x xy x xy F F FF F x F FF FF
0
" # " 0 # 0 0 0 0 0 0
1 ðA AF 0 CFÞ 1 ðQ =FÞ0 1 P0 P0 P P 1 E 1 0 ¼ H F 2H 3 F þ F þ F F þ þ F ; y yx2 y xy FF 00 FF 00 y2 F 00 F 00 F 00 F 00 F 00
0
ð2:10Þ
374
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
where
i 1 h 0 0 0 00 2ðA AF CFÞ ðA AF CFÞ F ; FF i 1 h 0 H ¼ 00 2ðB BF 0 DFÞ ðB BF 0 DFÞ F : FF G¼
These provide the constraints for the coefficient functions in Eq. (1.1)
A AF 0 CF ¼ 0;
Q ¼ c7 F 0 ; F
B BF 0 DF ¼ 0;
0 0 0 P P0 P P F F þ 00 ¼ 0; FF 00 FF F 00 F 00
E FF 00
0 F
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 ;
E0 ¼ c8 F 00 ;
P0 ¼ c9 F 00 ;
0 0 E E E0 þ 00 ¼ 0: 00 00 FF FF F
In this situation, we obtain that the solution of Eq. (1.1) can be given by
Z
u
pffiffiffiffiffi 1 ¼ ln jxyj þ c1 t þ c2 : FðzÞ
If F 00 ¼ 0, i.e., F ¼ u, then we also have the above formulas. m1 2k1 Let A ¼ B ¼ um ; F ¼ uk ; E ¼ F ¼ 0. We have C ¼ D ¼ umk ku and Q ¼ c7 ku . Thus we have shown that the equation
utt ¼ um ðuxx þ uyy Þ þ ðumk ku
m1
Þðu2x þ u2y Þ þ c7 ku
2k1
has the exact solution
( u¼
pffiffiffiffiffi c0 xy exp c1 t; k ¼ 1; 1 pffiffiffiffiffi ½ð1 kÞðln jxyj þ c1 t þ c2 Þk1 ; k – 1:
ð2:11Þ
Subcase 2.2 f ðxÞ ¼ ln j cos xj; gðyÞ ¼ ln j cos yj. Set f ðxÞ ¼ ln j cos xj; gðyÞ ¼ ln j cos yj. Then v x ¼ tan x; v y ¼ tan y. Moreover, we have E1 ¼ fu : ux ¼ tan xF; uy ¼ tan yFg. If F 00 – 0, then Eq. (1.3) becomes 02
h ¼
0 i 1 h Q 0 0 0 0 2 2 0 f ðA þ AF þ CFÞ F þ 2ðA þ AF þ CFÞ þ ðB þ BF þ DFÞ Fðtan yÞ þ A þ B þ F þ 2ðA þ AF 0 þ CFÞg ðtan xÞ F FF 00 E E0 E 1 P0 þ þ 00 tan y ð2:12Þ þ 00 00 tan x þ 00 FF FF tan x F F
and Eq. (1.4) becomes 02
h ¼
0 i 1 h Q 0 0 2 0 2 0 0 F þ 2ðB þ BF 0 þ DFÞg 00 f ðB þ BF þ DFÞ F þ 2ðB þ BF þ DFÞ ðtan yÞ þ ðA þ AF þ CFÞ Fðtan xÞ þ A þ B þ F FF P P0 P 1 E0 þ þ tan x: ð2:13Þ 00 þ 00 tan y þ F FF FF 00 tan y F 00
Differentiating Eq. (2.12) with respect to x and Eq. (2.13) with respect to y, we get
" # " # 0 0 0 0 0
ðB þ BF 0 þ DFÞ E E E E0 2 0 0 ¼ G0 F þ 2G ðtan xÞ3 þ F tan xðtan yÞ þ ð2G þ M FÞ tan x þ F þ þ F þ tan2 x F 00 F 00 F 00 FF 00 FF 00 0 0 0 E E0 P E 1 þ 00 F tan x tan y 00 þ F þ ; ð2:14Þ F 00 F FF 00 FF tan2 x " # 0 0
ðA þ AF 0 þ CFÞ 0 ¼ H0 F þ 2H ðtan yÞ3 þ F tan yðtan xÞ2 þ ð2H þ N0 FÞ tan y F 00 " # 0 0 0 0 0 0 P P P P0 P P0 E P 1 2 þ tan þ 00 F tan x tan y 00 F þ þ F þ y þ F þ ; FF 00 FF 00 FF 00 FF tan2 y F 00 F 00 F 00 F where
G¼
i 1 h 0 0 0 00 ðA þ AF þ CFÞ F þ 2ðA þ AF þ CFÞ ; FF
H¼
i 1 h 0 0 0 00 ðB þ BF þ DFÞ F þ 2ðB þ BF þ DFÞ ; FF
ð2:15Þ
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
M¼
1 FF 00
0 Q AþBþ F þ 2ðA þ AF 0 þ CFÞ ; F
N¼
1 FF 00
0 Q AþBþ F þ 2ðB þ BF 0 þ DFÞ : F
Solving them, we derive
A þ AF 0 þ CF ¼ 0;
E FF 00
0
Fþ
B þ BF 0 þ DF ¼ 0;
0 0 E E E0 F þ 00 ¼ 0; 00 þ 00 FF F F
AþBþ
P FF 00
0
Fþ
Q ¼ c10 F 0 ; F
m1
m1
pffiffiffiffiffi c1 t þ c2 ;
E0 ¼ c11 F 00 ;
P0 ¼ c12 F 00 ;
0 0 P P P0 F þ 00 ¼ 0: 00 þ 00 FF F F
Let A ¼ B ¼ um ; F ¼ uk ; E ¼ F ¼ 0. Then C ¼ D ¼ ðumk þ ku
utt ¼ um ðuxx þ uyy Þ ðumk þ ku
hðtÞ ¼
Þðu2x þ u2y Þ þ c10 ku
2k1
Þ and Q ¼ c10 ku
2k1
2umþk . Thus the equation
2umþk
has the exact solution
8 pffiffiffiffi < c0 exp ð c1 tÞ ; k ¼ 1; cos x cos y u¼ 1 pffiffiffiffiffi : ½ð1 kÞð ln j cos x cos yj þ c1 t þ c2 Þ1k ; k – 1: Subcase 2.3
2
2
f ðxÞ ¼ x2 ; gðyÞ ¼ y2 .
v x ¼ x; v y ¼ y and E1 ¼ fu : ux ¼ xF; uy ¼ yFg. If F 00 – 0, then Eq. (1.3) becomes
Taking f ðxÞ ¼ x2 =2; gðyÞ ¼ y2 =2. Then 02
h ¼
ð2:16Þ
1 h 0 0 ðAF 0 þ CFÞ Fx2 þ 2ðAF 0 þ CFÞ þ ðBF 0 þ DFÞ Fy2 FF 00 0 Q E 1 E0 P0 F þ 00 þ 00 x þ 00 y; þ AþBþ F FF x F F
ð2:17Þ
and Eq. (1.4) becomes 02
h ¼
0 1 Q P 1 E0 P0 0 0 0 0 2 0 2 ðBF þ DFÞ Fy þ 2ðBF þ DFÞ þ ðAF þ CFÞ Fx þ A þ B þ F þ 00 þ 00 x þ 00 y: 00 F FF FF y F F
ð2:18Þ
Differentiating Eq. (2.17) with respect to x and Eq. (2.18) with respect to y yields
" # " # ( 0 0 ) 0 0 0 0 0 0 ðAF 0 þ CFÞ ðBF 0 þ DFÞ 2ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ ðA þ B þ Q =FÞ0 E 1 E 3 2 xþ 0¼ x þ xy þ þ þ 2 2 00 00 00 00 00 00 x F F FF FF 00 F F FF F 0 0 0 0 E E0 P þ 00 x2 þ 00 þ 00 xy; F FF F " # " # ( 0 0 ) 0 0 0 0 0 0 0 ðBF 0 þ DFÞ ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ ðA þ B þ Q =FÞ0 E 1 P 3 2 y þ 00 xy 2 2 00 0¼ y þ yx þ þ þ 00 00 00 00 00 y F F FF F F FF F F 0 0 0 P P0 P þ þ 00 þ 00 y2 : 00 FF FF F Solving them, we obtain
AF 0 þ CF ¼ 0; E0 ¼ c14 F 00 ;
Q A þ B þ ¼ c13 F 0 ; F pffiffiffiffiffi hðtÞ ¼ c1 t þ c2 :
BF 0 þ DF ¼ 0;
P0 ¼ c15 F 00 ;
k1
Let A ¼ B ¼ um ; F ¼ uk . Then E ¼ c14 ku m1
utt ¼ um ðuxx þ uyy Þ ku
; P ¼ c15 ku
ðu2x þ u2y Þ þ c14 ku
k1
k1
; C ¼ D ¼ ku k1
ux þ c15 ku
m1
2k1
and Q ¼ c13 ku 2k1
uy þ c13 ku
2umþk . Thus the equation
2umþk
has the exact solution
u¼
8 ffi < c0 exp x2 þy2 pffiffiffiffi c1 t ; k ¼ 1; 2 :
½ð1 kÞð ln j cos x cos yj þ
1 pffiffiffiffiffi c1 t þ c2 Þ1k ; k – 1:
ð2:19Þ
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
In summary, we utilize the method of invariant set to derive exact solutions of a class of wave equation. It is noted that we can also apply this approach to solve higher dimensional wave equation. Acknowledgments The authors would like to express their sincere thanks to the anonymous referees for their valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
V.A. Galaktionov, Groups of scaling and invariant sets for higher-order nonlinear evolution equations, Differ. Integral Equ. 14 (2001) 913–924. V.A. Galaktionov, Ordered invariants sets for nonlinear evolution equations of KDV-type, Comput. Math. Math. Phys. 39 (1999) 1564–1570. C.Z. Qu, P.G. Estevez, Extended rotation and reduction of evolution equations, Nonlinear Anal. 52 (2003) 1655–1673. C.Z. Qu, Symmetries and solutions to the thin film equations, J. Math. Anal. Appl. 317 (2006) 381–397. C.Z. Qu, C.R. Zhu, Invariant sets and solutions to the generalized thin film equation, Sci. China Ser. A 50 (2007) 875–886. C.R. Zhu, C.Z. Qu, Invariant sets and solutions to higher-dimensional reaction–diffusion equations with source term, Phys. Lett. A 354 (2006) 437–444. L. Martina, M.B. Sheftel, P. Winternitz, Group foliation and non-invariant solutions of the heavenly equation, J. Phys. A 34 (2001) 9243–9263. R.Z. Zhdanov, Conditional Lie-Bücklund symmetry and reduction of evolution equation, J. Phys. A 28 (1995) 3841–3850. G.W. Bluman, J.D. Cole, The general similarity solution of the heat equation, J. Math. Fluid Mech. 18 (1969) 1025–1042. G.W. Bluman, S. Kumei, Symmetries and Differential Equations, Springer-Verlag, New York, 1989. C.Z. Qu, S.L. Zhang, R.C. Liu, Separation of variables and exact solutions to quasilinear diffusion equations with nonlinear source, Physica D 144 (2000) 97–123. S.I. Betelü, J.R. King, Explicit solutions of a two-dimensional fourth-order nonlinear diffusion equation, Math. Comput. Model. 37 (2003) 395–403. J.R. King, Exact polynomial solutions to some nonlinear diffusion equations, Physica D 64 (1993) 35–65. G.Z. Qu, S.L. Zhang, C.R. Zhu, Invariant sets and exact solutions to higher-dimensional wave equations, Commun. Theor. Phys. 49 (2008) 1119–1124. R. Cherniha, M. Henkel, On nonlinear partial differential equations with an infinite-dimensional conditional symmetry, J. Math. Anal. Appl. 298 (2004) 487–500.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Invariant sets and solutions to a class of wave equations q Yaying Dong a,⇑, Shunli Zhang c, Shanbing Li b a b c
School of Mathematics, Northwest University, Xi’an 710069, PR China College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China Center for Nonlinear Studies, School of Mathematics, Northwest University, Xi’an 710069, PR China
a r t i c l e
i n f o
a b s t r a c t In this paper, we investigate the invariant sets and exact solutions of the (1 + 2)-dimensional wave equations. It is proven that there exists a class of solutions to the equations, which belong to the invariant set E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Wave equation Invariant set Exact solution
1. Introduction In mathematical physics, seeking exact solutions of nonlinear partial differential equations (PDEs) has been the highlights. Constructing exact solutions to nonlinear PDEs has theoretical significance and practical value in physics, engineering and mechanics, etc. Up to now, the scholars have developed various ways to solve nonlinear systems. In [1], Galaktionov introduced an extension to the scaling group for the equation
ut ¼ Kðx; ux ; uxx ; . . . ; uðkÞ Þ: It is governed by the invariance of the set S0 ¼ fu : ux ¼ FðuÞ=xg, where uðkÞ represents the kth-order derivative of u with respect to x and F is a smooth function to be determined. This approach is available to solve the nonlinear evolution equations of the KdV-type[2], and it has also been used to study the second and fourth-order nonlinear evolution equations [1]. In [3], Qu and Estevez extended the scaling group to a more general form, which is governed by the invariant set
S1 ¼
u : ux ¼
Z 1 FðuÞ þ FðuÞ expðn 1Þ x
u
1 dz : FðzÞ
This method has been used successfully to construct exact solutions to a large amount of evolution equations (see [3,4]). Moreover, in [5,6], Qu and Zhu further extended the Galaktionov’s method to study the generalized thin film equation and the (1 + 2)-dimensional reaction–diffusion equations by introducing the invariant set E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg. There have been many works contributed to constructing exact solutions of the nonlinear diffusion equation (see [7–9]). Symmetry-related methods are valid to construct exact solutions of nonlinear diffusion equations. Local symmetry-related methods including nonclassical symmetry method due to Bluman and Cole [9], Lie point symmetry method due to Lie [10], and conditional Lie Bäcklund symmetry method due to Zhdanov [8]. Another important approach is the ansatz-based method, which has been used to obtain exact solutions of the following equation
ut ¼ DGðuÞ þ HðuÞ; q
The project is supported by the National Natural Science Foundation of China (No. 11371293).
⇑ Corresponding author.
E-mail address: [email protected] (Y. Dong). http://dx.doi.org/10.1016/j.amc.2014.12.093 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
even for higher-dimensional case. It is noted that the symmetry-related methods have been used successfully to study the symmetry properties and the solutions of nonlinear PDEs (see [11–15]). In present paper, we extend Galaktionov’s approach [1,2] to study a class of wave equations and their solutions, which are governed by the invariant set E0 . Note that the invariant set is a natural generalization of S0 in [1] for the (1 + 1)-dimensional case. Consider the (1 + 2)-dimensional wave equation
utt ¼ AðuÞuxx þ BðuÞuyy þ CðuÞu2x þ DðuÞu2y þ EðuÞux þ PðuÞuy þ Q ðuÞ;
ð1:1Þ
where AðuÞ; BðuÞ; CðuÞ; DðuÞ; EðuÞ; PðuÞ and Q ðuÞ are smooth functions of u. Eq. (1.1) has a wide range of physical applications in fluid dynamics, plasma and elastic media, etc. Now, we introduce the invariant set E0 which is governed by
E0 ¼ fu : ux ¼ v x FðuÞ; uy ¼ v y FðuÞg; where v ðx; yÞ is a smooth function of x and y; FðuÞ is a smooth function of u, which is to be determined from the invariant condition
uðx; y; 0Þ 2 E0 ) uðx; y; tÞ 2 E0 ; t 2 ð0; 1: For u 2 E0 , we can obtain the solutions of Eq. (1.1), which are given by
Z
u
1 dz ¼ v ðx; yÞ þ hðtÞ: FðzÞ
In the invariant set E0 , we derive the following formulas 0
00
02
ut ¼ h F;
utt ¼ h F þ h FF 0 ;
ux ¼ v x F;
uxx ¼ v xx F þ v 2x FF 0 ;
uy ¼ v y F;
uyy ¼ v yy F þ v 2y FF 0 :
Substituting them into Eq. (1.1), we have 00
02
h þ h F 0 ¼ Av xx þ Bv yy þ ðAF 0 þ CFÞv 2x þ ðBF 0 þ DFÞv 2y þ Ev x þ Pv y þ
Q : F
ð1:2Þ
Differentiating (1.2) with respect to x and y respectively, we have
0 Q 02 h F 00 v x F ¼ Av xxx þ Bv yyx þ A0 v xx þ B0 v yy þ v x F þ ðAF 0 þ CFÞ0 F v 3x F 0
þ 2ðAF 0 þ CFÞv x v xx þ ðBF 0 þ DFÞ v x F v 2y þ 2ðBF 0 þ DFÞv y v yx
ð1:3Þ
þ Ev xx þ E0 v 2x F þ P0 v x v y F þ Pv yx ; 0 Q 02 h F 00 v y F ¼ Av xxy þ Bv yyy þ A0 v xx þ B0 v yy þ E0 v x þ P0 v y þ v yF F 0
0
þ ðAF 0 þ CFÞ F v y v 2x þ 2ðAF 0 þ CFÞv x v xy þ ðBF 0 þ DFÞ F v 3y
ð1:4Þ
þ 2ðBF 0 þ DFÞv y v yy þ Ev xy þ Pv yy : By means of Eqs. (1.3) and (1.4), it is very difficult to determine the coefficient functions in Eq. (1.1). To illustrate our method, in Section 2, we only choose certain v ðx; yÞ and calculate the exact solutions of Eq. (1.1). 2. Main results In this section, we choose certain v ðx; yÞ and construct the corresponding solutions of Eq. (1.1) to determine the invariant sets. However, we cannot determine all v ðx; yÞ such that Eq. (1.1) is invariant with respect to E0 , we can consider several special cases in the following. Case 1 v xx þ v yy ¼ 0. With respect to the solutions of the Laplace’s equation
v xx þ v yy ¼ 0; we get the following subcases.
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
Subcase 1.1 Substituting 02
h ¼
371
v ðx; yÞ ¼ ax þ by ða – 0; b – 0Þ. v ðx; yÞ ¼ ax þ by into Eq. (1.2), we obtain
1 2 Q 2 00 0 0 : a ðAF þ CFÞ þ b ðBF þ DFÞ þ aE þ bP þ h F F0
ð2:1Þ
It is noted that the left side of Eq. (2.1) does not depend on x; y, which suggests that A; B; C; D; E; P and Q satisfy
1 2 Q 2 00 0 0 ¼ c1 a ðAF þ CFÞ þ b ðBF þ DFÞ þ aE þ bP þ h F F0 and hðtÞ ¼ solution
Z
u
ð2:2Þ
pffiffiffiffiffi c1 t þ c2 , where c1 P 0 and ci denote arbitrary constants from now on. In this subcase, Eq. (1.1) has the exact
pffiffiffiffiffi 1 dz ¼ ax þ by þ c1 t þ c2 ; FðzÞ
which is the traveling wave solution. Subcase 1.2 v ¼ arctanðx=yÞ. The function v ¼ arctanðx=yÞ is also a solution of Laplace equation. In this situation, E0 becomes the set
E1 ¼
y FðuÞ; u : ux ¼ 2 x þ y2
uy ¼
x FðuÞ : x2 þ y2
If F 00 – 0, then Eqs. (1.3) and (1.4) respectively become
h
02
¼
i
1 x3 1 h x2 0 0 þ 00 6ðA BÞ þ ðBF 0 þ DFÞ F 00 2ðBF þ DFÞ 2 2 FF FF yðx2 þ y2 Þ ðx2 þ y2 Þ
1 xy 2ðA BÞ0 F þ 4ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 2 2 FF 00 ðx þ y2 Þ
i 1 h y2 ðQ =FÞ0 0 þ 00 2ðA BÞ þ ðAF 0 þ CFÞ F þ 2 2 2 FF F 00 ðx þ y Þ 2E P0 x E0 P y P x2 þ 00 00 2 þ 00 ; 00 þ 00 2 2 2 2 FF FF x þ y FF yðx þ y2 Þ F x þy F
h
02
¼
i
1 y3 1 h y2 0 0 00 6ðA BÞ ðAF 0 þ CFÞ F 00 2ðAF þ CFÞ 2 2 FF FF xðx2 þ y2 Þ ðx2 þ y2 Þ
1 xy 2ðA BÞ0 F 4ðBF 0 þ DFÞ þ 2ðAF 0 þ CFÞ 2 FF 00 ðx2 þ y2 Þ
i 1 h x2 ðQ=FÞ0 0 þ 00 2ðA BÞ þ ðBF 0 þ DFÞ F þ 2 FF F 00 ðx2 þ y2 Þ 0 0 E 2P y P E x E y2 þ 00 00 2 þ þ : 00 00 FF x þ y2 FF x2 þ y2 FF 00 xðx2 þ y2 Þ F F Differentiating Eqs. (2.3) and (2.4) with respect to x and y respectively, we obtain
0 ¼
ð2:3Þ
" 0 0 #
1 x4 BF þ DF 2x3 0 2ðBF þ DFÞ G þ F 00 00 2 2 2 2 2 2 FF FF y ðx þ y Þ yðx þ y2 Þ 6ðBF 0 þ DFÞ x2 2xy þ G0 F þ 6M þ ðG M0 F 2IÞ 00 2 2 2 2 2 FF ðx þ y2 Þ ðx þ y Þ " # 0 0 0 0 y2 ðQ =FÞ0 E 2E0 P 4P x þ ðI0 F 2MÞ þ F þ 2 F F þ 2 FF 00 FF 00 x2 þ y2 F 00 F 00 F 00 ðx2 þ y2 Þ " 0 0 0 # 2E P 0 P x2 E0 2E P0 P y þ þ þ F F F 2 ; þ x þ y2 yðx2 þ y2 Þ FF 00 F 00 FF 00 FF 00 F 00 FF 00 F 00
ð2:4Þ
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
0¼
" 0 0 #
1 y4 AF þ CF 2y3 0 2ðAF þ CFÞ H F 00 00 2 2 FF FF x2 ðx2 þ y2 Þ xðx2 þ y2 Þ
6ðAF 0 þ CFÞ y2 2xy H0 F þ þ 6N þ ðH N 0 F þ 2JÞ 00 2 2 FF ðx2 þ y2 Þ ðx2 þ y2 Þ " # 0 0 0 0 x2 ðQ=FÞ0 P 2P0 E 4E y 0 þ ðJ F þ 2NÞ þ F þ 2 F 00 þ 00 F 00 2 2 F 00 F F FF 00 FF x þ y2 ðx2 þ y2 Þ " 0 0 0 # 2P E0 E y2 P0 2P E0 E x þ 00 þ 00 þ F F 2 ; þ 00 F þ 00 00 x þ y2 xðx2 þ y2 Þ FF FF 00 FF FF 00 F F F
where
1 0 ½6ðA BÞ þ ðBF 0 þ DFÞ F; FF 00
H¼
1 0 ½6ðA BÞ ðAF 0 þ CFÞ F; FF 00
1 0 ½2ðA BÞ þ ðAF 0 þ CFÞ F; FF 00
J¼
1 0 ½2ðA BÞ þ ðBF 0 þ DFÞ F; FF 00
G¼
I¼
M¼
1 ½ðA BÞ0 F þ 2ðAF 0 þ CFÞ ðBF 0 þ DFÞ; FF 00
N¼
1 ½ðA BÞ0 F 2ðBF 0 þ DFÞ þ ðAF 0 þ CFÞ: FF 00
They provide the constraints for the coefficient functions in Eq. (1.1)
AF 0 þ CF ¼ 0;
A ¼ B;
BF 0 þ DF ¼ 0;
Q ¼ c3 F 0 ; F
0 2E P0 P þ þ F ¼ 0; FF 00 F 00 FF 00
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 ;
2
E FF 00
0
F
0 0 2E0 P 4P F þ 00 ¼ 0; FF F 00 F 00
0 0 0 E 2E P0 P F F ¼ 0: F 00 FF 00 F 00 FF 00
In this situation, Eq. (1.1) has the exact solution
Z
u
pffiffiffiffiffi 1 x þ c1 t þ c2 : dz ¼ arctan FðzÞ y
If F 00 ¼ 0, i.e., F ¼ u, similarly, we also have the above formulas. It is easy to verify that the equation m1
utt ¼ um ðuxx þ uyy Þ ku
ðu2x þ u2y Þ þ c3 ku
2k1
has the exact solution
pffiffiffiffiffi 8 > < c0 exp arctan yx þ c1 t ; k ¼ 1; u¼ h i 1 ffi > k1 : ð1 kÞ arctan x þ pffiffiffiffi ; k – 1: c1 t þ c2 y Subcase 1.3 v xx ¼ v yy ¼ 0. Since v xx ¼ v yy ¼ 0, we derive that becomes the set
v ðx; yÞ ¼ cxy þ d.
ð2:5Þ
Without loss of generality, we choose c ¼ 1 and d ¼ 0. Thus E0
E1 ¼ fu : ux ¼ yFðuÞ; uy ¼ xFðuÞg: If F 00 – 0, substituting
v xx ¼ v yy ¼ 0 into Eqs. (1.3) and (1.4) respectively, then we get
0 1 2ðBF 0 þ DFÞ x Q 1 0 0 02 h ¼ 00 ðBF 0 þ DFÞ x2 þ ðAF 0 þ CFÞ y2 þ þ E0 y þ P0 x þ þ P ; F y F Fy F
ð2:6Þ
0 1 2ðAF 0 þ CFÞ y Q 1 0 2 0 2 0 0 0 0 : E ðBF þ DFÞ x þ ðAF þ CFÞ y þ þ E y þ P x þ þ F x F Fx F 00
ð2:7Þ
02
h ¼
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
373
Differentiating Eqs. (2.6) and (2.7) with respect to x and y respectively, we obtain
" # ( ) " # 0 0 0 0 0 0 0 ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 1 ðQ=FÞ0 2 xþ 0¼ Fx y þ Fþ Fy3 þ þ Fy 00 00 00 00 00 00 y FF FF F F F F þ
0 0 0 0 0 E P P0 P 2 Fy þ Fyx þ þ F; FF 00 F 00 F 00 F 00
" # ( ) " # 0 0 0 0 0 0 0 ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ ðBF 0 þ DFÞ 2ðAF 0 þ CFÞ 1 ðQ =FÞ0 2 3 y þ Fxy þ F þ Fx þ þ Fx x F 00 F 00 F 00 F 00 FF 00 FF 00 0 0 0 0 0 E P E0 E F: þ 00 Fyx þ 00 Fx2 þ 00 þ F F F FF 00
0¼
Hence, we derive the coefficient functions in Eq. (1.1) satisfying
AF 0 þ CF ¼ 0;
BF 0 þ DF ¼ 0;
Q ¼ c4 F 0 ; F
E0 ¼ c5 F 00 ;
P 0 ¼ c6 F 00
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 :
In this situation, Eq. (1.1) has the exact solution
Z
u
pffiffiffiffiffi 1 dz ¼ xy þ d þ c1 t þ c2 : FðzÞ
If F 00 ¼ 0, i.e., F ¼ u, we also have the above formulas. k1 k1 m1 2k1 Let A ¼ B ¼ um ; F ¼ uk . We have E ¼ c5 ku ; P ¼ c6 ku ; C ¼ D ¼ ku and Q ¼ c4 ku . Thus we have shown that the equation m1
utt ¼ um ðuxx þ uyy Þ ku
k1
ðu2x þ u2y Þ þ c5 ku
ux þ c6 ku
k1
2k1
uy þ c4 ku
has the exact solution
( u¼
pffiffiffiffiffi c1 tÞ; k ¼ 1; 1 pffiffiffiffiffi ½ð1 kÞðxy þ c1 t þ c2 Þk1 ; k – 1:
c0 expðxy þ
ð2:8Þ
Case 2 v xy ¼ 0. From v xy ¼ 0, we deduce v ðx; yÞ ¼ f ðxÞ þ gðyÞ, where f ðxÞ and gðyÞ are the smooth functions of x and y respectively. In this case, E0 becomes 0
E1 ¼ fu : ux ¼ f ðxÞF; uy ¼ g 0 ðyÞFg: We now consider several subcases. Subcase 2.1 f ðxÞ ¼ ln jxj; gðyÞ ¼ ln jyj. 0 Set f ðxÞ ¼ ln jxj; gðyÞ ¼ ln jyj. Then f ðxÞ ¼ 1=x;
1 1 E1 ¼ u : ux ¼ F; uy ¼ F : x y
g 0 ðyÞ ¼ 1=y; v ðx; yÞ ¼ ln jxj þ ln jyj. Moreover, we have
Suppose that F 00 – 0. Then Eq. (1.3) becomes 02
h ¼
0 i 1 1 1 1h 1 Q E E0 1 P0 1 0 0 0 0 0 2ðA AF CFÞ ðA AF CFÞ F ðB BF DFÞ þ þ FF 00 x2 F 00 y2 F 00 F FF 00 F 00 x F 00 y
ð2:9Þ
and Eq. (1.4) becomes 02
h ¼
0 0 i 1 1 1 1h 1 Q E0 1 P P 1 0 0 0 0 0 2ðB BF DFÞ ðB BF DFÞ F ðA AF CFÞ þ þ þ : F 00 x F 00 FF 00 y FF 00 y2 F 00 x2 F 00 F
Differentiating Eqs. (2.9) and (2.10) with respect to x and y respectively, we have
" # " # 0 0 0 0 0 0 0 0
1 ðB BF 0 DFÞ 1 ðQ=FÞ0 1 E E E E0 1 P 1 0 ¼ G F 2G 3 F 2þ F F 00 þ 00 2 þ 00 F ; 00 00 00 00 x xy x xy F F FF F x F FF FF
0
" # " 0 # 0 0 0 0 0 0
1 ðA AF 0 CFÞ 1 ðQ =FÞ0 1 P0 P0 P P 1 E 1 0 ¼ H F 2H 3 F þ F þ F F þ þ F ; y yx2 y xy FF 00 FF 00 y2 F 00 F 00 F 00 F 00 F 00
0
ð2:10Þ
374
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
where
i 1 h 0 0 0 00 2ðA AF CFÞ ðA AF CFÞ F ; FF i 1 h 0 H ¼ 00 2ðB BF 0 DFÞ ðB BF 0 DFÞ F : FF G¼
These provide the constraints for the coefficient functions in Eq. (1.1)
A AF 0 CF ¼ 0;
Q ¼ c7 F 0 ; F
B BF 0 DF ¼ 0;
0 0 0 P P0 P P F F þ 00 ¼ 0; FF 00 FF F 00 F 00
E FF 00
0 F
hðtÞ ¼
pffiffiffiffiffi c1 t þ c2 ;
E0 ¼ c8 F 00 ;
P0 ¼ c9 F 00 ;
0 0 E E E0 þ 00 ¼ 0: 00 00 FF FF F
In this situation, we obtain that the solution of Eq. (1.1) can be given by
Z
u
pffiffiffiffiffi 1 ¼ ln jxyj þ c1 t þ c2 : FðzÞ
If F 00 ¼ 0, i.e., F ¼ u, then we also have the above formulas. m1 2k1 Let A ¼ B ¼ um ; F ¼ uk ; E ¼ F ¼ 0. We have C ¼ D ¼ umk ku and Q ¼ c7 ku . Thus we have shown that the equation
utt ¼ um ðuxx þ uyy Þ þ ðumk ku
m1
Þðu2x þ u2y Þ þ c7 ku
2k1
has the exact solution
( u¼
pffiffiffiffiffi c0 xy exp c1 t; k ¼ 1; 1 pffiffiffiffiffi ½ð1 kÞðln jxyj þ c1 t þ c2 Þk1 ; k – 1:
ð2:11Þ
Subcase 2.2 f ðxÞ ¼ ln j cos xj; gðyÞ ¼ ln j cos yj. Set f ðxÞ ¼ ln j cos xj; gðyÞ ¼ ln j cos yj. Then v x ¼ tan x; v y ¼ tan y. Moreover, we have E1 ¼ fu : ux ¼ tan xF; uy ¼ tan yFg. If F 00 – 0, then Eq. (1.3) becomes 02
h ¼
0 i 1 h Q 0 0 0 0 2 2 0 f ðA þ AF þ CFÞ F þ 2ðA þ AF þ CFÞ þ ðB þ BF þ DFÞ Fðtan yÞ þ A þ B þ F þ 2ðA þ AF 0 þ CFÞg ðtan xÞ F FF 00 E E0 E 1 P0 þ þ 00 tan y ð2:12Þ þ 00 00 tan x þ 00 FF FF tan x F F
and Eq. (1.4) becomes 02
h ¼
0 i 1 h Q 0 0 2 0 2 0 0 F þ 2ðB þ BF 0 þ DFÞg 00 f ðB þ BF þ DFÞ F þ 2ðB þ BF þ DFÞ ðtan yÞ þ ðA þ AF þ CFÞ Fðtan xÞ þ A þ B þ F FF P P0 P 1 E0 þ þ tan x: ð2:13Þ 00 þ 00 tan y þ F FF FF 00 tan y F 00
Differentiating Eq. (2.12) with respect to x and Eq. (2.13) with respect to y, we get
" # " # 0 0 0 0 0
ðB þ BF 0 þ DFÞ E E E E0 2 0 0 ¼ G0 F þ 2G ðtan xÞ3 þ F tan xðtan yÞ þ ð2G þ M FÞ tan x þ F þ þ F þ tan2 x F 00 F 00 F 00 FF 00 FF 00 0 0 0 E E0 P E 1 þ 00 F tan x tan y 00 þ F þ ; ð2:14Þ F 00 F FF 00 FF tan2 x " # 0 0
ðA þ AF 0 þ CFÞ 0 ¼ H0 F þ 2H ðtan yÞ3 þ F tan yðtan xÞ2 þ ð2H þ N0 FÞ tan y F 00 " # 0 0 0 0 0 0 P P P P0 P P0 E P 1 2 þ tan þ 00 F tan x tan y 00 F þ þ F þ y þ F þ ; FF 00 FF 00 FF 00 FF tan2 y F 00 F 00 F 00 F where
G¼
i 1 h 0 0 0 00 ðA þ AF þ CFÞ F þ 2ðA þ AF þ CFÞ ; FF
H¼
i 1 h 0 0 0 00 ðB þ BF þ DFÞ F þ 2ðB þ BF þ DFÞ ; FF
ð2:15Þ
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Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
M¼
1 FF 00
0 Q AþBþ F þ 2ðA þ AF 0 þ CFÞ ; F
N¼
1 FF 00
0 Q AþBþ F þ 2ðB þ BF 0 þ DFÞ : F
Solving them, we derive
A þ AF 0 þ CF ¼ 0;
E FF 00
0
Fþ
B þ BF 0 þ DF ¼ 0;
0 0 E E E0 F þ 00 ¼ 0; 00 þ 00 FF F F
AþBþ
P FF 00
0
Fþ
Q ¼ c10 F 0 ; F
m1
m1
pffiffiffiffiffi c1 t þ c2 ;
E0 ¼ c11 F 00 ;
P0 ¼ c12 F 00 ;
0 0 P P P0 F þ 00 ¼ 0: 00 þ 00 FF F F
Let A ¼ B ¼ um ; F ¼ uk ; E ¼ F ¼ 0. Then C ¼ D ¼ ðumk þ ku
utt ¼ um ðuxx þ uyy Þ ðumk þ ku
hðtÞ ¼
Þðu2x þ u2y Þ þ c10 ku
2k1
Þ and Q ¼ c10 ku
2k1
2umþk . Thus the equation
2umþk
has the exact solution
8 pffiffiffiffi < c0 exp ð c1 tÞ ; k ¼ 1; cos x cos y u¼ 1 pffiffiffiffiffi : ½ð1 kÞð ln j cos x cos yj þ c1 t þ c2 Þ1k ; k – 1: Subcase 2.3
2
2
f ðxÞ ¼ x2 ; gðyÞ ¼ y2 .
v x ¼ x; v y ¼ y and E1 ¼ fu : ux ¼ xF; uy ¼ yFg. If F 00 – 0, then Eq. (1.3) becomes
Taking f ðxÞ ¼ x2 =2; gðyÞ ¼ y2 =2. Then 02
h ¼
ð2:16Þ
1 h 0 0 ðAF 0 þ CFÞ Fx2 þ 2ðAF 0 þ CFÞ þ ðBF 0 þ DFÞ Fy2 FF 00 0 Q E 1 E0 P0 F þ 00 þ 00 x þ 00 y; þ AþBþ F FF x F F
ð2:17Þ
and Eq. (1.4) becomes 02
h ¼
0 1 Q P 1 E0 P0 0 0 0 0 2 0 2 ðBF þ DFÞ Fy þ 2ðBF þ DFÞ þ ðAF þ CFÞ Fx þ A þ B þ F þ 00 þ 00 x þ 00 y: 00 F FF FF y F F
ð2:18Þ
Differentiating Eq. (2.17) with respect to x and Eq. (2.18) with respect to y yields
" # " # ( 0 0 ) 0 0 0 0 0 0 ðAF 0 þ CFÞ ðBF 0 þ DFÞ 2ðAF 0 þ CFÞ 2ðAF 0 þ CFÞ ðA þ B þ Q =FÞ0 E 1 E 3 2 xþ 0¼ x þ xy þ þ þ 2 2 00 00 00 00 00 00 x F F FF FF 00 F F FF F 0 0 0 0 E E0 P þ 00 x2 þ 00 þ 00 xy; F FF F " # " # ( 0 0 ) 0 0 0 0 0 0 0 ðBF 0 þ DFÞ ðAF 0 þ CFÞ 2ðBF 0 þ DFÞ 2ðBF 0 þ DFÞ ðA þ B þ Q =FÞ0 E 1 P 3 2 y þ 00 xy 2 2 00 0¼ y þ yx þ þ þ 00 00 00 00 00 y F F FF F F FF F F 0 0 0 P P0 P þ þ 00 þ 00 y2 : 00 FF FF F Solving them, we obtain
AF 0 þ CF ¼ 0; E0 ¼ c14 F 00 ;
Q A þ B þ ¼ c13 F 0 ; F pffiffiffiffiffi hðtÞ ¼ c1 t þ c2 :
BF 0 þ DF ¼ 0;
P0 ¼ c15 F 00 ;
k1
Let A ¼ B ¼ um ; F ¼ uk . Then E ¼ c14 ku m1
utt ¼ um ðuxx þ uyy Þ ku
; P ¼ c15 ku
ðu2x þ u2y Þ þ c14 ku
k1
k1
; C ¼ D ¼ ku k1
ux þ c15 ku
m1
2k1
and Q ¼ c13 ku 2k1
uy þ c13 ku
2umþk . Thus the equation
2umþk
has the exact solution
u¼
8 ffi < c0 exp x2 þy2 pffiffiffiffi c1 t ; k ¼ 1; 2 :
½ð1 kÞð ln j cos x cos yj þ
1 pffiffiffiffiffi c1 t þ c2 Þ1k ; k – 1:
ð2:19Þ
376
Y. Dong et al. / Applied Mathematics and Computation 253 (2015) 369–376
In summary, we utilize the method of invariant set to derive exact solutions of a class of wave equation. It is noted that we can also apply this approach to solve higher dimensional wave equation. Acknowledgments The authors would like to express their sincere thanks to the anonymous referees for their valuable suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
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